所以，我试图用python制作一个基本的井字游戏，然后我创建了一个效果很好的游戏，但是我的代码并不好，因为它有很多代码可以检查列表索引（游戏胜利者），这让我很烦。那么，如何避免使用左索引来检查游戏的获胜者？
我的代码：
board = [' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ']
def show_board():# for showing the tic-tac-toe board
print(' | ' + str(board[0]) + ' | ' +
str(board[1]) + ' | ' + str(board[2]) + ' | ')
print(' | ' + str(board[3]) + ' | ' +
str(board[4]) + ' | ' + str(board[5]) + ' | ')
print(' | ' + str(board[6]) + ' | ' +
str(board[7]) + ' | ' + str(board[8]) + ' | ')
def main():
one = 1
flag = 1
show_board()
while one == 1:
if flag == 1:
x_o = 'X'
if flag == 2:
x_o = 'O'
pos = int(input('Player "' + x_o + '" Turn: '))
if x_o == 'o':
x_o = 'O'
if x_o == 'x':
x_o = 'X'
if board[pos - 1] == 'O' or board[pos - 1] == 'O':
print('That Place Is Already Filled By Player "0"')
if board[pos - 1] == 'X' or board[pos - 1] == 'X':
print('That Place Is Already Filled By Player "X"')
else:
try:
board[pos - 1] = x_o
except IndexError:
print('Type Numbers Between Only 1 And 9')
if flag == 1:
flag = 2
elif flag == 2:
flag = 1
show_board()
# Checking The Winner Of The Game
# for horizontal X
if board[0] == board[1] == board[2] == 'X':
one = 2
print('The Winner Is Player "X"!')
elif board[3] == board[4] == board[5] == 'X':
one = 2
print('The Winner Is Player "X"!')
elif board[6] == board[7] == board[8] == 'X':
one = 2
print('The Winner Is Player "X"!')
# for Daigonal X
elif board[0] == board[4] == board[8] == 'X':
one = 2
print('The Winner Is Player "X"!')
elif board[2] == board[4] == board[6] == 'X':
one = 2
print('The Winner Is Player "X"!')
# for Vertical X
elif board[1] == board[4] == board[7] == 'X':
one = 2
print('The Winner Is Player "X"!')
elif board[2] == board[5] == board[8] == 'X':
one = 2
print('The Winner Is Player "X"!')
elif board[0] == board[3] == board[6] == 'X':
one = 2
print('The Winner Is Player "X"!')
# for horizontal O
elif board[0] == board[1] == board[2] == 'O':
one = 2
print('The Winner Is Player "O"!')
elif board[3] == board[4] == board[5] == 'O':
one = 2
print('The Winner Is Player "O"!')
elif board[6] == board[7] == board[8] == 'O':
one = 2
print('The Winner Is Player "O"!')
# for Diagonal O
elif board[0] == board[4] == board[8] == 'O':
one = 2
print('The Winner Is Player "O"!')
elif board[2] == board[4] == board[6] == 'O':
one = 2
print('The Winner Is Player "O"!')
# for Vertical 0
elif board[1] == board[4] == board[7] == 'O':
one = 2
print('The Winner Is Player "O"!')
elif board[2] == board[5] == board[8] == 'O':
one = 2
print('The Winner Is Player "O"!')
elif board[0] == board[3] == board[6] == 'O':
one = 2
print('The Winner Is Player "O"!')
elif board[0] != ' ' and board[1] != ' ' and board[2] != ' ' and board[3] != ' ' and board[4] != ' ' and board[5] != ' ' and board[6] != ' ' and board[7] != ' ' and board[8] != ' ':
print('The Match Is A Tie!')
one = 2
main()
因此，如您所见，我使用了大量的if语句来检查游戏的获胜者。如何避免这种情况并减少行数。
答案 0 :(得分：3)
您可以将它们替换为一些循环，例如：
board = [' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ']
def show_board():# for showing the tic-tac-toe board
print(' | ' + str(board[0]) + ' | ' +
str(board[1]) + ' | ' + str(board[2]) + ' | ')
print(' | ' + str(board[3]) + ' | ' +
str(board[4]) + ' | ' + str(board[5]) + ' | ')
print(' | ' + str(board[6]) + ' | ' +
str(board[7]) + ' | ' + str(board[8]) + ' | ')
def main():
one = 1
flag = 1
show_board()
while one == 1:
if flag == 1:
x_o = 'X'
if flag == 2:
x_o = 'O'
pos = int(input('Player "' + x_o + '" Turn: '))
if x_o == 'o':
x_o = 'O'
if x_o == 'x':
x_o = 'X'
if board[pos - 1] == 'O' or board[pos - 1] == 'O':
print('That Place Is Already Filled By Player "0"')
if board[pos - 1] == 'X' or board[pos - 1] == 'X':
print('That Place Is Already Filled By Player "X"')
else:
try:
board[pos - 1] = x_o
except IndexError:
print('Type Numbers Between Only 1 And 9')
if flag == 1:
flag = 2
elif flag == 2:
flag = 1
show_board()
# Checking The Winner Of The Game
won = False
for turn in ('X', 'O'):
# horizontal
if not won:
for i in (0, 3, 6):
if all(board[i + k] == turn for k in range(3)):
won = True
break
# vertical
if not won:
for i in range(3):
if all(board[i + k] == turn for k in (0, 3, 6)):
won = True
break
# diagonal
if not won:
if all(board[k] == turn for k in (0, 4, 8)) or \
all(board[k] == turn for k in (2, 4, 6)):
won = True
# handle winning
if won:
one = 2
print(f'The Winner Is Player "{turn}"!')
break
# handle a tie
if not won and all(square != ' ' for square in board):
one = 2
print('The Match Is A Tie!')
main()
（您可能还可以对其他代码进行很多改进/简化）
以下是一些类似的代码，简化/完善/扩展为：
NUM_ROWS = 3
NUM_COLS = 3
NUM_WIN = 3
BOARD_SIZE = NUM_ROWS * NUM_COLS
EMPTY = ' '
BOARD = [EMPTY] * BOARD_SIZE
TURNS = 'X', 'O'
def show_board(board):
"""Show the tic-tac-toe board."""
for i in range(0, BOARD_SIZE, NUM_COLS):
print(' | ' + ' | '.join(board[i:i + NUM_COLS]) + ' | ')
def ij(i, j):
"""Convert (row, col) to board index."""
return i + NUM_COLS * j
def check_winner(board, turn):
"""Check if there is a winner."""
# horizontal
for i in range(NUM_ROWS):
for j in range(NUM_COLS - NUM_WIN + 1):
if all(board[ij(i, j + k)] == turn for k in range(NUM_WIN)):
return True
# vertical
for i in range(NUM_ROWS - NUM_WIN + 1):
for j in range(NUM_COLS):
if all(board[ij(i + k, j)] == turn for k in range(NUM_WIN)):
return True
# diagonal
for i in range(NUM_ROWS - NUM_WIN + 1):
for j in range(NUM_COLS - NUM_WIN + 1):
K = NUM_WIN
if all(board[ij(i + k, j + k)] == turn for k in range(NUM_WIN)):
return True
if all(board[ij(i + NUM_WIN - k - 1, j + k)] == turn
for k in range(NUM_WIN)):
return True
def check_tie(board):
"""Check if tie."""
return all(square != EMPTY for square in board)
def next_turn(turn):
"""Advance to next turn."""
return TURNS[(TURNS.index(turn) + 1) % 2]
def main():
"""Tic-tac-toe game."""
turn = TURNS[0]
show_board(BOARD)
while True:
valid_input = False
while not valid_input:
try:
choice = int(input(f'Player `{turn}` turn: '))
valid_input = (1 <= choice <= BOARD_SIZE)
if not valid_input:
raise ValueError
except ValueError:
print(f'Type numbers between 1 and {BOARD_SIZE} only.')
else:
idx = choice - 1
if BOARD[idx] != EMPTY:
print(f'Position `{idx}` already taken by `{BOARD[idx]}`')
else:
BOARD[idx] = turn
show_board(BOARD)
won = check_winner(BOARD, turn)
if won:
print(f'The winner is player `{turn}`!')
break
# handle a tie
if not won and check_tie(BOARD):
print('The match is a tie!')
break
turn = next_turn(turn)
if __name__ == '__main__':
main()
请使用NumPy和类（尤其是Board.py）检查mnk-game的source code以获得类似的实现。免责声明：我是包裹的作者。
请注意，仅对于井字游戏，可以使用类似于@ rusu_ro1 proposed的方法来缩短check_winner()：
WIN_CASES = [
[0, 1, 2], [3, 4, 5], [6, 7, 8], # horizontal
[0, 3, 6], [1, 4, 7], [2, 5, 8], # vertical
[0, 4, 8], [2, 4, 6], # diagonal
]
def check_winner(board, turn, win_cases=WIN_CASES):
"""Check if there is a winner."""
for win_case in win_cases:
if all(board[i] == turn for i in win_case):
return True
答案 1 :(得分：1)
import random as rnd
size = 3
## random board to test
board = [[rnd.choice(["X", "O"]) for _ in range(size)] for _ in range(size)]
print(board)
def check_winner(board):
for mark in ["X", "O"]:
## count mark in all rows
count_in_rows = [row.count(mark) for row in board]
## count mark in all columns
count_in_columns = [ [board[row_n][col_n] for row_n in range(size)].count(mark) for col_n in range(size)]
## count mark in fwd diagonal
count_in_diag = [[board[i][i] for i in range(size)].count(mark)]
## count in backward diagonal
count_in_diag.append([board[i][size-i-1] for i in range(size)].count(mark))
print((count_in_rows + count_in_columns + count_in_diag))
## if there is 3 count of mark in any rows, columns or diagonal
if 3 in (count_in_rows + count_in_columns + [count_in_diag]):
print(f"winner is {mark}")
return mark
return
check_winner(board)
答案 2 :(得分：1)
您可以定义o_list = ['o'，'o'，'o']和x_list = ['x'，'x'，'x'] 然后像这样比较它们：
#l = board
if(l[0:3] == o_list or l[3:6] == o_list or l[6:9] == o_list or l[0:9:3] == o_list or l[1:9:3] == o_list or l[8:0:-3] == o_list):
print('The Winner Is Player "O"!')
与x_list相同。 ;）
o_list =['o', 'o', 'o']
x_list =['x', 'x', 'x']
board = [' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ']
def show_board():# for showing the tic-tac-toe board
print(' | ' + str(board[0]) + ' | ' +
str(board[1]) + ' | ' + str(board[2]) + ' | ')
print(' | ' + str(board[3]) + ' | ' +
str(board[4]) + ' | ' + str(board[5]) + ' | ')
print(' | ' + str(board[6]) + ' | ' +
str(board[7]) + ' | ' + str(board[8]) + ' | ')
def main():
one = 1
flag = 1
o_list =['o', 'o', 'o']
x_list =['x', 'x', 'x']
show_board()
while one == 1:
if flag == 1:
x_o = 'X'
if flag == 2:
x_o = 'O'
pos = int(input('Player "' + x_o + '" Turn: '))
if x_o == 'o':
x_o = 'O'
if x_o == 'x':
x_o = 'X'
if board[pos - 1] == 'O' or board[pos - 1] == 'O':
print('That Place Is Already Filled By Player "0"')
if board[pos - 1] == 'X' or board[pos - 1] == 'X':
print('That Place Is Already Filled By Player "X"')
else:
try:
board[pos - 1] = x_o
except IndexError:
print('Type Numbers Between Only 1 And 9')
if flag == 1:
flag = 2
elif flag == 2:
flag = 1
show_board()
# Checking The Winner Of The Game
# for horizontal X
if(board[0:3] == x_list or board[3:6] == x_list or board[6:9] == x_list or board[0:9:3] == x_list or board[1:9:3] == x_list or board[8:0:-3] == x_list):
one = 2
print('The Winner Is Player "X"!')
if(board[0:3] == o_list or board[3:6] == o_list or board[6:9] == o_list or board[0:9:3] == o_list or board[1:9:3] == o_list or board[8:0:-3] == o_list):
one = 2
print('The Winner Is Player "O"!')
elif board[0] != ' ' and board[1] != ' ' and board[2] != ' ' and board[3] != ' ' and board[4] != ' ' and board[5] != ' ' and board[6] != ' ' and board[7] != ' ' and board[8] != ' ':
print('The Match Is A Tie!')
one = 2
main()
答案 3 :(得分：0)
您可以使用以下功能（输入：板子和玩家的字母，输出：如果该玩家赢了，则为True）：
def isWinner(board, letter):
return (
(board[1] == board[2] == board[3] == letter) or
(board[4] == board[5] == board[6] == letter) or
(board[7] == board[8] == board[9] == letter) or
(board[7] == board[4] == board[1] == letter) or
(board[8] == board[5] == board[2] == letter) or
(board[9] == board[6] == board[3] == letter) or
(board[7] == board[5] == board[3] == letter) or
(board[9] == board[5] == board[1] == letter))
答案 4 :(得分：0)
要检查获胜者，您可以使用：
from operator import itemgetter
from itertools import chain
winning_cases = {'diags' : [[0, 4, 8], [2, 4, 6]],
'rows': [[0, 1, 2], [3, 4, 5], [6, 7, 8]],
'cols': [[0, 3, 6], [1, 4, 7], [2, 5, 8]]}
def check_winner(board=board):
for case in chain(*winning_cases.values()):
candidates = set(itemgetter(*case)(board))
if len(candidates) == 1 and '' not in candidates: # check if a diag, row or col has only one charcter
winner, = candidates
return winner
return None
例如：
board = ['X', 'X', 'O', 'X', 'O','O', 'X', 'O', 'X']
check_winner(board)
输出：
X
如果有一个，check_winner方法将返回赢家，否则返回无